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प्रश्न
Find the equation of the plane which passes through the point (3, 4, –1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes
उत्तर
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 .........(1)
2x – 3y + 5z + λ = 0 ..........(2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(–1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 ........(3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2
b = – 3
c = 5
d1 = 7
d2 = 11
Distance = `|("d"_1 - "d"_2)/sqrt("a"^2 + "b"^2 + "c"^2)|`
= `|(7 - 11)/sqrt(2^2 + 3^2 + 5^2)|`
= `|4/sqrt(38)|`
d = `4/sqrt(38)`
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संबंधित प्रश्न
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Choose the correct alternative:
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